并发动作和线程 [英] Concurrent actions and threads
本文介绍了并发动作和线程的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我想同时执行两个名为A和B的方法.
I want to concurrently execute two methods, named A and B.
我也希望B等待A完成.
I also want B to wait for A to finish.
如何通过用Java实现线程来获得这样的结果?
How can I achieve such results by implementing threads in Java?
推荐答案
Use Thread#join(). Call it on thread object which death you want to wait for.
join方法允许一个线程等待另一个线程的完成.
The join method allows one thread to wait for the completion of another.
官方教程中的示例:
public class SimpleThreads {
// Display a message, preceded by
// the name of the current thread
static void threadMessage(String message) {
String threadName =
Thread.currentThread().getName();
System.out.format("%s: %s%n",
threadName,
message);
}
private static class MessageLoop
implements Runnable {
public void run() {
String importantInfo[] = {
"Mares eat oats",
"Does eat oats",
"Little lambs eat ivy",
"A kid will eat ivy too"
};
try {
for (int i = 0;
i < importantInfo.length;
i++) {
// Pause for 4 seconds
Thread.sleep(4000);
// Print a message
threadMessage(importantInfo[i]);
}
} catch (InterruptedException e) {
threadMessage("I wasn't done!");
}
}
}
public static void main(String args[])
throws InterruptedException {
// Delay, in milliseconds before
// we interrupt MessageLoop
// thread (default one hour).
long patience = 1000 * 60 * 60;
// If command line argument
// present, gives patience
// in seconds.
if (args.length > 0) {
try {
patience = Long.parseLong(args[0]) * 1000;
} catch (NumberFormatException e) {
System.err.println("Argument must be an integer.");
System.exit(1);
}
}
threadMessage("Starting MessageLoop thread");
long startTime = System.currentTimeMillis();
Thread t = new Thread(new MessageLoop());
t.start();
threadMessage("Waiting for MessageLoop thread to finish");
// loop until MessageLoop
// thread exits
while (t.isAlive()) {
threadMessage("Still waiting...");
// Wait maximum of 1 second
// for MessageLoop thread
// to finish.
t.join(1000);
if (((System.currentTimeMillis() - startTime) > patience)
&& t.isAlive()) {
threadMessage("Tired of waiting!");
t.interrupt();
// Shouldn't be long now
// -- wait indefinitely
t.join();
}
}
threadMessage("Finally!");
}
}
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